Tính tổng
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\)
\(B=\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{2015}{2016}\)
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2017}}\)
chứng tỏ A<1
2,
\(S=2+2^2+2^3+...+2^{99}\)
C/t: S chia hết cho 7, 31
3,
\(A=1+5+5^2+5^3+5^4+5^5+...+5^{99}+5^{100}\)
Tính A
4,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)<1
5,
CHỨNG tỏ rằng các p/s tối giản vs mọi số tự nhiên n
a,\(\frac{n+1}{2n+3}\)b,\(\frac{2n+3}{4n+8}\)
6,
a,TÍnh A và B
A=\(\frac{2016^{2016}+2}{2016^{1016}-1}\)B=\(\frac{2016^{2016}}{2016^{2016}-3}\)
b, tính
C=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
LÀm NHANH Hộ MK ,MAi mk Phải Nộp.
Tính tổng
B=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2016}}\)
Các bạn làm đầy đủ giùm mình nha!!!
(NHANH+ĐÚNG=LIKE cho câu trả lời)
Tính
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(C=\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{51}}-\frac{1}{5^{52}}\)
bài 1
a, \(\left(-\frac{2}{5}\right):\left[\frac{-3}{5}+\frac{3}{2}\right]\)
b, \(\frac{-5}{2016}.\frac{2}{11}+\frac{-5}{2016}.\frac{9}{11}+1\frac{5}{2016}\)
Tính:
\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}.\)
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}.\)
Cho biểu thức sau: \(P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+.....+\frac{2015}{5^{2015}}+\frac{2016}{5^{2016}}\)
Chứng minh 1/4 < P< 1/3
Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
Gửi bạn ... nè :
B=1/1*3+1/3*5/+1/5*7+....+1/2015*2016
\(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2016}\)
\(B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2016}\right)\)
\(B=\frac{1}{2}.\frac{2015}{2016}\)
\(B=\frac{2015}{4032}\)